uln2803

Let's Go! · 2010-02-03 15:57

what is the minimum voltage required to operate the uln 2803?·for testing purposes, can one put +v directly from the battery to the input pin to get an output, or does it have to be ttl or cmos input?

thanks

▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
The smarter I get, the more I understand I don't know!

kf4ixm · 2010-02-03 16:02

According to the datasheet, it can handle up to 30v on the input. switching from off to on at about 2.4 to 2.7v

http://www.sparkfun.com/datasheets/IC/uln2803a.pdf
·

Let's Go! · 2010-02-03 16:12

that's what i thought. i can't even get them to sx on an led using a 9 volt battery. any ideas?

▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
The smarter I get, the more I understand I don't know!

Let's Go! · 2010-02-03 16:14

do they require v regulators?

▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
The smarter I get, the more I understand I don't know!

Let's Go! · 2010-02-03 16:22

does the output pin go high up to almost supply voltage to actuate the relay?

▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
The smarter I get, the more I understand I don't know!

kf4ixm · 2010-02-03 16:24

they do not require regulators in order to operate, they basically bring the outputs to gnd or common line potential (sinking output) with an input of +2.7v or greater on the input side. try using an ohm meter between the common and output, (black lead·on common, red lead on output)·then apply voltage to the input. you should then see the ohm meter go to 0 ohm or there abouts.

Let's Go! · 2010-02-03 16:36

thanks, i'll have done that and you just confirmed my understanding.

▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
The smarter I get, the more I understand I don't know!

Let's Go! · 2010-02-03 16:37

so would i connect the relay coil directly to the output pin to drive the relay?

▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
The smarter I get, the more I understand I don't know!

kf4ixm · 2010-02-03 16:43

yes, one side of the coil would go to +v the other side would go to the output pin of the uln2803. it actually doesn't 'drive' the relay,(just my point of view being that drive in a device·produces or 'source'·a + voltage)·as the uln2803 doesn't source, (provides a path to + voltage), it's a sinking device, (provides a path to gnd or common).

Let's Go! · 2010-02-03 16:48

ok, so if i connect one side of the relay coil to +v and the other side to the 2803 ouput leading to the relay coil , when the 2803 input actuates the 2803 output, the relay will actuate with + v sinking to the 2803 which is now ground level?

is that the way it works, mostly?

▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
The smarter I get, the more I understand I don't know!

Let's Go! · 2010-02-03 16:50

thanks, i had it backwards as i was using the 2803 as a source. i really appreciate your help. now i need to change my boards. jim

▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
The smarter I get, the more I understand I don't know!

kf4ixm · 2010-02-03 16:52

Yep, you got it!

Let's Go! said...
ok, so if i connect one side of the relay coil to +v and the other side to the 2803 ouput leading to the relay coil , when the 2803 input actuates the 2803 output, the relay will actuate with + v sinking to the 2803 which is now ground level?

is that the way it works, mostly?

Let's Go! · 2010-02-03 17:04

thanks, i wouldn't have figured that out in a month of sundays. jim

▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
The smarter I get, the more I understand I don't know!

allanlane5 · 2010-02-03 18:22

If you're using this to drive relays, make sure you hook pin 10 up to the + voltage as well. This connects a set of protection diodes for each output up to +V.

A relay needs this, because a magnetic field is generated when you turn it on. The collapsing of this magnetic field when you turn it off generates a·current, which needs some place to go. Connecting pin 10 to the +V gives this current some place to go. Otherwise, it will burn out the ULN2803's output transistor.

uln2803

Comments